Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $x = \dfrac{p + 4}{p + 9} \div \dfrac{p^2 - 6p - 40}{-3p + 30} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{p + 4}{p + 9} \times \dfrac{-3p + 30}{p^2 - 6p - 40} $ First factor the quadratic. $x = \dfrac{p + 4}{p + 9} \times \dfrac{-3p + 30}{(p + 4)(p - 10)} $ Then factor out any other terms. $x = \dfrac{p + 4}{p + 9} \times \dfrac{-3(p - 10)}{(p + 4)(p - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (p + 4) \times -3(p - 10) } { (p + 9) \times (p + 4)(p - 10) } $ $x = \dfrac{ -3(p + 4)(p - 10)}{ (p + 9)(p + 4)(p - 10)} $ Notice that $(p - 10)$ and $(p + 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -3\cancel{(p + 4)}(p - 10)}{ (p + 9)\cancel{(p + 4)}(p - 10)} $ We are dividing by $p + 4$ , so $p + 4 \neq 0$ Therefore, $p \neq -4$ $x = \dfrac{ -3\cancel{(p + 4)}\cancel{(p - 10)}}{ (p + 9)\cancel{(p + 4)}\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $x = \dfrac{-3}{p + 9} ; \space p \neq -4 ; \space p \neq 10 $